$F/K$ algebraic and every nonconstant polynomial in $K[X]$ has a root in $F$ implies $F$ is algebraically closed.

Let $E/F$ be a finite extension (which is separable as the characteristic is zero), we will prove that $E=F$, hence $F$ is algebraically closed: Let $a$ be a primitive element of $E/F$. Then $a$ is algebraic over $K$, so there exists a finite Galois extension $N/K$ containing $a$, in particular $$ E\subseteq NF. $$ (One can take $N$ to be the Galois closure of $K(a)$, for example.) Let $f\in K[x]$ be the minimal polynomial of a primitive element of $N$ over $K$. Since $N/K$ is Galois, any of the roots of $f$ generates $N$ over $K$.

Now, by assumption $f$ has a root in $F$, so $N\subseteq F$, hence $E\subseteq NF=F$. This proves that $F$ has no non-trivial extensions, hence $F$ is algebraically closed.